If $f$ is an entire function of order $\lambda$ then $f'$ also has order $\lambda$

If $f$ is an entire function of order $\lambda$ then $f'$ also has order $\lambda$

Can someone show me how to prove this or point me in the right direction? My definition is an entire function $f$ has finite order $\lambda$ if for $\epsilon>0$ we have $|f(z)| < $ exp$(|z|^{\lambda + \epsilon})$ for all $|z|$ sufficiently large. I also have that $\lambda = $ inf$\{ a : |f(z)| < $ exp$(|z|^{a}) $ for $|z|$ sufficiently large}, though I'm not sure if that will be useful here. I had to show the order of the sum of two functions is less than or equal to the max of the order of each function in an earlier problem, but I'm not sure how to go about this for $f'$.

Thanks for any help.

Solution 1:

Denote with $$ M(r, f) = \max\{ |f(z)| : |z| \le r \} $$ the maximum of $|f|$ on disks of radius $r$. It follows from Cauchy's integral formula for the derivative that $$ M(r, f') \le M(r+1, f) \, , $$ which implies $\lambda(f') \le \lambda(f)$. (For the details, see for example $f$ is entire, show that the order of $f'$ is less than or equal to that of $f$ .)

For the other direction note that $$ f(r e^{i \phi}) = f(0) + \int_0^r f'(t e^{i \phi})e^{i \phi} \, dt $$ implies $$ M(r, f) \le |f(0)| + r M(r, f') \, . $$ from which $\lambda(f) \le \lambda(f')$ follows in a similar way.